NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions

In this exercise 3.4 of the trigonometric function, we will study the solutions of trigonometric equations. The equation containing trigonometric functions are known as trigonometric equations,like sinx = siny, cosx +1 =0 etc. The solutions of these trigonometric equations are of two kinds.

1-Principal Solution. The values of an unknown angle between 0° and 360°that satisfy the given equation are known as principal solutions of the given equation.

2-General Solutions. Set of infinite solutions that satisfy the given equation is known as the general solution, this solution is lead due to the periodicity of the trigonometric functions. The general solution depends upon period of the trigonometric function, as an example if principal solution of the equation sinx =1 ⇒ sinx= sinπ/2,so principal solution is x =π/2 then, this equation will satisfy for x= (2π ±π/2), (4π± π/2)…..etc, so in compact form, we can write these solutions x =nπ +(–1)n×π/2, it is all about the general solution. The periodicity and general solutions of trigonometric functions are given below.

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions

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Q1. Find the principal and general solution of the equation tanx = √3.

Answer.

${\color{DarkBlue} \mathbf{tanx = \sqrt{3}}}$

${\color{DarkBlue} \mathbf{{\color{DarkBlue} \mathbf{\sqrt{3}= tan\frac{\Pi }{3}}},tan\left (\Pi + \frac{\Pi }{3} \right )=tan\frac{4\Pi }{3}}}$

Hint: All possible values of x lying between 0 and 2π which satisfy the equation are called principal solutions.

Therefore, substituting √3 by tanπ/3 and tan 4π/3 we get

${\color{DarkBlue} \mathbf{tanx = tan\frac{\Pi }{3}\; and \; tanx= tan\frac{4\Pi }{3}}}$

${\color{DarkBlue} \mathbf{x =\frac{\Pi }{3}\; and\; x =\frac{4\Pi }{3}}}$

${\color{DarkBlue} \mathbf{\therefore The \; principal\; solutions\; will \; be\frac{\Pi }{3}\; and\; \frac{4\Pi }{3}}}$

For calculating general solution if we have tanx = tany then the value of x = nπ + y is known as general solution of the equation tanx = tany,so

Taking the smaller value of the solution π/3

${\color{DarkBlue} \mathbf{tanx =tan\frac{\Pi }{3}}}$

${\color{DarkBlue} \mathbf{x =n\Pi +\frac{\Pi }{3}}}$

4π/3 was not taken here because it is one of the infinite solutions ( for n=1, x = π +π/3 =4π/3)

${\color{DarkBlue} \mathbf{Therefore \; x=n\Pi +\frac{\Pi }{3}}}$ (n∈ z)

will be the general solution of the equation.

Q2. Find the principal and general solution of the equation secx = 2.

Answer.

secx =2

${\color{DarkBlue} \mathbf{2 = sec\frac{\Pi }{3}}}$

${\color{DarkBlue}\mathbf{secx=} \mathbf{sec\frac{\Pi }{3}}}$

${\color{DarkBlue}\mathbf{x=} \mathbf{\frac{\Pi }{3}}}$

Since the value of secx is positive(i.e 2),x will lie on 1 st and 4 th quadrant.(value of secx is positive in 1st and 4 th quadrant).

4 th quadrant means 2π −x (x =π/3)

${\color{DarkBlue}\mathbf{secx=} \mathbf{sec(2\Pi -\frac{\Pi }{3})}}$

${\color{DarkBlue}\mathbf{secx=} \mathbf{sec\frac{5\Pi }{3}}}$

${\color{DarkBlue}\mathbf{x=} \mathbf{\frac{5\Pi }{3}}}$

Therefore principal solutions of the equation will be

${\color{DarkBlue} \mathbf{\frac{\Pi }{3}\; and\; \frac{5\Pi }{3}}}$

If secx = secy then general solution of the equation is given by the same expression as for the cosx =cosy, see the hint.

Hint: If cosx = cosy⇒1/secx = 1 /secy⇒secx=secy,therefore secx =secy implies that cosx = cosy.

${\color{DarkBlue} \mathbf{secx = sec\frac{\Pi }{3}}}$

${\color{DarkBlue} \mathbf{cosx = cos\frac{\Pi }{3}}}$

${\color{DarkBlue} \mathbf{x = 2n\Pi \pm \frac{\Pi }{3}}}$

(n ∈ z)

Therefore the general solution of the equation will be 2nπ ± π/3.

Q3. Find the principal and general solutions of the equation cotx = –√3.

Answer.

cotx = –√3.

√3 = cotπ/6

${\color{DarkBlue} \mathbf{cotx = -cot\frac{\Pi }{6}}}$

Value of cotx is negative ,so x will lie on 2nd quadrant (i.e π– π/6) and fourth quadrant(i.e 2π– π/6)(value of cotx is negative in 2 nd and 4 th quadrant).

${\color{DarkBlue} \mathbf{cotx=cot\left ( \Pi -\frac{\Pi }{6} \right )=cot\left ( 2\Pi -\frac{\Pi }{6} \right )}}$

${\color{DarkBlue} \mathbf{cotx=cot\frac{5\Pi }{6}=cot\frac{11\Pi }{6}}}$

Therefore the principal solutions of the given equation will be,

${\color{DarkBlue} \mathbf{\frac{5\Pi }{6}\; and\; \frac{11\Pi }{6}.}}$

${\color{DarkBlue} \mathbf{cotx=cot\frac{5\Pi }{6}} }$

${\color{DarkBlue} \mathbf{x= 2n\Pi +\frac{5\Pi }{6}} }$

Therefore the general solution of the given equation is

${\color{DarkBlue} \mathbf{2n\Pi +\frac{5\Pi }{6}.} }$

Q4.Find the general solution of cosecx =–2.

Answer.

${\color{DarkBlue} \mathbf{cosecx= -cosec\frac{\Pi }{6}}}$

Value of cosec is negative in 3rd(π+π/6) and 4 th quadrant (2π–π/6).

${\color{DarkBlue} \mathbf{cosecx=cosec(\Pi +\frac{\Pi }{6})= cosec(2\Pi -\frac{\Pi }{6})}}$

${\color{DarkBlue} \mathbf{cosecx=cosec\frac{7\Pi }{6}= cosec\frac{11\Pi }{6}}}$

Therefore the principal solutions of the given equation will be,

${\color{DarkBlue} \mathbf{\frac{7\Pi }{6}\; and\; \frac{11\Pi }{6}.}}$

The general solution of the given equation will be

${\color{DarkBlue} \mathbf{cosecx = cosec\frac{7\Pi }{6}}}$

As we know cosecx = 1/sinx

${\color{DarkBlue} \mathbf{sinx =sin\frac{7\Pi }{6}}}$

${\color{DarkBlue} \mathbf{x= n\Pi +\left ( -1 \right )^{n}\frac{7\Pi }{6},\; \; \; n\epsilon z}}$

Q5.Find the general solution of the equation cos4x = cos2x.

Answer.

cos4x = cos2x

cos4x – cos2x = 0

${\color{DarkBlue} \mathbf{\because cosA - cosB =2sin\frac{A+B}{2}sin\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 2sin\frac{4x+2x}{2}sin\frac{4x-2x}{2}=0}}$

⇒2sin3x.sinx= 0

⇒sin3x=0 and sinx =0

sin3x=0

sin3x = sin0

${{\mathbf{3x=\left n\Pi +( -1 \right )^{n}\Theta }}}$

since θ = 0

3x = nπ

${\color{DarkBlue} \mathbf{x =\frac{n\Pi }{3}}}$

sinx =0 =sin0

x = nπ

Therefore general solution of both the equation will be

${\color{DarkBlue} \mathbf{n\Pi \; or\; \frac{n\Pi }{3}.}}$

Q6. Find the general solution of the equation cos3x + cosx – cos2x = 0.

Answer.

cos3x + cosx – cos2x = 0

(cos3x + cosx )– cos2x = 0

${\color{DarkBlue} \mathbf{\because cosA + cosB = 2cos\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 2cos\frac{3x+x}{2}.cos\frac{3x-x}{2}-cos2x=0}}$

⇒2cos2x.cosx – cos2x =0

⇒cos2x(2cosx – 1) = 0

Since general solution of the equation cosx = cosy is x = 2nπ ± y,so

cos2x = 0

${\color{DarkBlue} \mathbf{cos2x=cos\frac{\Pi }{2}}}$

${\color{DarkBlue} \mathbf{2x=2n\Pi \pm \frac{\Pi }{2}}}$

${\color{DarkBlue} \mathbf{x=n\Pi \pm \frac{\Pi }{4}}}$

2cosx – 1 =0

2cosx = 1

${\color{DarkBlue} \mathbf{cosx =\frac{1}{2}}}$

${\color{DarkBlue} \mathbf{cosx = cos\frac{\Pi }{3}}}$

${\color{DarkBlue} \mathbf{x = 2n\Pi \pm \frac{\Pi }{3}}}$

Therefore the general solution of the given equation will be

${\color{DarkBlue} \mathbf{n\Pi \pm \frac{\Pi }{4} \; and \; 2n\Pi \pm \frac{\Pi }{3}.}}$

Q7. Find the general solution of the equation sin2x + cosx = 0.

Answer.

sin2x + cosx = 0

2sinx.cosx + cosx = 0

cosx(2sinx + 1) = 0

cosx = 0, 2sinx + 1 = 0

cosx = cos0, sinx = –1/2⇒sinx = -sinπ/6 = sin(π+ π/6)=sin7π/6

${\color{DarkBlue} \mathbf{x =2n\Pi \pm \frac{\Pi }{2} \; or\; sinx=sin\frac{7\Pi }{6}\Rightarrow x =n\Pi\pm \left (-1 \right )^{n}\frac{7\Pi }{6}}}$

Therefore the general solution of the given equation will be ,

${\color{DarkBlue} \mathbf{2n\Pi \pm \frac{\Pi }{2}\; and\; \; n\Pi \pm \left ( -1 \right )^{n}\frac{7\Pi }{6}}}$

Q8. Find the general solution of the equation sec²2x = 1 – tan2x.

Answer.

sec²2x = 1 – tan2x

1 + tan²2x = 1 – tan2x

tan²2x + tan2x = 0

tan2x(tan2x +1) = 0

tan2x = 0

tan2x = tan0( general solution of the equation (tanx = tany) is x = nπ + y)

2x = nπ + 0

${\color{DarkBlue} \mathbf{x=\frac{n\Pi }{2}}}$

tan2x +1 = 0

tan2x = –1

${\color{DarkBlue} \mathbf{tan2x = -tan\frac{\Pi }{4}}}$

Value of tan is negative so 2x will lye in 2nd(π –π/4) and 4 th quadrant (2π – π/4)[tanθ is negative in 2nd and 4 th quadrant].

${\color{DarkBlue} \mathbf{tan2x= tan\left ( \Pi -\frac{\Pi }{4} \right )=tan\left ( 2\Pi -\frac{\Pi }{4} \right )}}$

Taking the smaller value of 2x

${\color{DarkBlue} \mathbf{tan2x =\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{2x =n\Pi +\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{x =\frac{n\Pi }{2}+\frac{3\Pi }{8}}}$

Therefore the general solution of the given equation is

${\color{DarkBlue} \mathbf{\frac{n\Pi }{2}\; and\; \frac{n\Pi }{2}+\frac{3\Pi }{8}.}}$ (n εz)

Q9. Find the general solution of the equation sinx + sin3x + sin5x = 0.

Answer.

sinx + sin3x + sin5x = 0

(sinx + sin5x )+ sin3x = 0

${\color{DarkBlue} \mathbf{\because sinA +sinB = 2sin\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 2sin\frac{x+5x}{2}.cos\frac{x-5x}{2}+sin3x}}$

⇒2sin3x.cos(–4x) + sin3x=0

⇒2sin3x.cos4x +sin3x=0

⇒sin3x(2cos4x +1) =0

sin3x = 0 and 2cos4x +1 =0⇒cos4x = –1/2=–cosπ/3

${\color{DarkBlue} \mathbf{3x =n\Pi +\left ( -1 \right )^{n}\times 0}}$

${\color{DarkBlue} \mathbf{x =\frac{n\Pi }{3}}}$

cos4x = –cosπ/3

Value of cos4x is negative in 2nd (π -π/3) and 3rd quadrant(π + π/3),taking smaller value in getting general solution.

${\color{DarkBlue} \mathbf{\Rightarrow cos4x =cos(\Pi -\frac{\Pi }{3})}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos4x =cos\frac{2\Pi }{3}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 4x =2n\Pi \pm \frac{2\Pi }{3}}}$

${\color{DarkBlue} \mathbf{\Rightarrow x =\frac{n\Pi }{2} \pm \frac{\Pi }{6}}}$

Therefore the general solution of the given equation is

${\color{DarkBlue} \mathbf{\frac{n\Pi }{3}\; and\; \frac{n\Pi }{2} \pm \frac{\Pi }{6}.}}$

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
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NCERT Solutions for class 10 maths

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Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

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NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions

CBSE Class 11-Question paper of maths 2015

NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions